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anonymous

  • 5 years ago

Find all the values of x in the interval [0, 2pie] that satisfy the equation 2cos(x) + sin(2x) = 0.

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  1. anonymous
    • 5 years ago
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    step one is to rewrite \[sin(2x)=2sin(x)cos(x)\]

  2. anonymous
    • 5 years ago
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    then you get \[2cos(x)+2cos(x)sin(x)=0\] \[2cos(x)(1+sin(x))=0\] \[cos(x)=0\] or \[1+sin(x)=0\] \[sin(x)=-\frac{1}{2}\]

  3. anonymous
    • 5 years ago
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    can you solve from there?

  4. anonymous
    • 5 years ago
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    can you continue please?

  5. anonymous
    • 5 years ago
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    sure. you are in the interval \[(0,2\pi)\]

  6. anonymous
    • 5 years ago
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    in that interval cosine is 0 at \[\frac{\pi}{2}\] and \[\frac{3\pi}{2}\]

  7. anonymous
    • 5 years ago
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    are those fractions?

  8. anonymous
    • 5 years ago
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    if you do not instantly know where \[sin(x)=-\frac{1}{2}\] then look at the cheat sheet http://tutorial.math.lamar.edu/cheat_table.aspx and see that it is at \[\frac{7\pi}{6}\] and \[\frac{11\pi}{6}\]

  9. anonymous
    • 5 years ago
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    that is where the second coordinate is \[-\frac{1}{2}\]

  10. anonymous
    • 5 years ago
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    unit circle on last page of cheat sheet

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